Leetcode 42. Trapping Rain Water

最新推荐文章于 2023-08-07 09:17:41 发布

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本文介绍了一个算法，用于计算给定高度数组所代表的地形在雨后能够储存的雨水量。通过实现一个类，我们能够针对任意长度的非负整数数组进行计算，并返回雨后储存的雨水总量。

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

public class Solution {
    public int trap(int[] height) {
      int length = height.length;

      int maxL = new int[length];
      int maxR = new int[length];

      if(n <= 2)
        return 0;

      int max = 0;
      for(int i = 1; i < n - 1; i ++){
        maxL[i] = max;
        if(max < height[i])
          max = A[i];
      }
      max = A[n - 1]
      maxR[n - 1] = 0;
      int ctrap,ttrap = 0;
      for(int i = n-2; i > 0; i--){
        maxR[i] = max;
        ctra[]
      }
    }
}